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Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left Therefore, the pH will be higher than the pH of the acid solution lowering the H 3 O + ion concentration

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Calculate the pH of a solution which is 1.0 M HF and 1.0M NaF. Note: Last chapter we calculated pH of 1.0 M solution of HF only. We calculated the pH to be 1.57.

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Buffers Buffers: Solutions that resist changes in pH when a small amount of acid or base is added. They act by neutralizing the added acid or base Many buffers are made by mixing:  A weak acid with its conjugate base anion HF + NaF   acidic buffer  A weak base with it’s conjugate acid cation NH 3 + NH 4 +   basic buffer

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Calculate the pH of a buffer which is 0.50M in acetic acid (CH 3 CO 2 H) and 0.50M in sodium acetate (NaCH 3 CO 2 ). Calculate the pH after 0.01 mol HCl is added to one liter of the above buffer. Compare with the pH of one liter of water after 0.01 mol of HCl is added to it.

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Henderson-Hasselbalch Equation Calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson- Hasselbalch Equation The equation calculates the pH of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base Note: the “x is small” approximation must be valid

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Deriving the Henderson-Hasselbalch Equation ()()()()()()

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Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? The Henderson-Hasselbalch equation is generally good enough when the “x is small” (conjugate acid/base initial concentrations high, K a small) Rule of thumb: the initial acid and salt concentrations should be over 1000x larger than the value of K a

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Calculate the pH of a buffer solution containing 0.25 M NH 3 and 0.40 M NH 4 Cl. K b (NH 3 )= 1.8 x10 -5 Then, if we add 20mL of 1.0M NaOH to 500mL of the above buffer, what would the final pH be? Note: the volume of the buffer changes in this problem, so we need to work in moles and calculate new concentrations.